Question: For $-25 \le x \le 25,$ find the maximum value of $\sqrt{25 + x} + \sqrt{25 - x}.$
Explanation: By QM-AM,
\[\frac{\sqrt{25 + x} + \sqrt{25 - x}}{2} \le \sqrt{\frac{25 + x + 25 - x}{2}} = 5,\]so $\sqrt{25 + x} + \sqrt{25 - x} \le 10.$

Equality occurs at $x = 0,$ so the maximum value is $\boxed{10}.$